Full-Length Accuplacer Math Practice Test-Answers and Explanations
34- Choice C is correct.
If the score of Mia was 72, then Ava’s score of Ava is 36. Since the score of Emma was two-thirds of that Ava’s, therefore, the score of Emma is 24.
35- Choice A is correct.
Let \(x\) be the number. Write the equation and solve for \(x\).
\(\frac{3}{7} ×21= \frac{3}{5}. x ⇒ (\frac{3×21}{7}= \frac{3x}{5}\), use cross multiplication to solve for \(x\).
\(5×63=3x×7 ⇒315=21x ⇒ x=15\)
36- Choice C is correct.
Let \(x\) be the number of years. Therefore, $1,200 per year equals \(1200x\). Starting from a $15,000 annual salary means you should add that amount to \(1200x\).
Income more than that is: \(I>1200x+15000\)
37- Choice D is correct.
average \(=\frac{sum \ of \ terms}{number \ of \ terms}\), The sum of the weight of all girls is: 15×50=750 kg
The sum of the weight of all boys is: 25×60=1500 kg
The sum of the weight of all students is: 750+1500=2250 kg, average \(=\frac{2250}{40}=56.25\)
38- Choice D is correct.
Write the numbers in order: 6,10,13,15,18,21,25
Since we have 7 numbers (7 is odd), the median is the number in the middle, which is 15.
39- Choice A is correct.
Formula for the Surface area of a cylinder is: \(SA=2πr^2+2πrh→48π=2πr^2+2πr(5)→r^2+5r-24=0 ⇒ (r+8)(r-3)=0→r=3 or r= -8\) (unacceptable)
40- Choice B is correct.
Let \(x\) be the smallest number. Then, these are the numbers: \(x, x+1, x+2, x+3, x+4, x+5\). Average \(= \frac{sum \ of \ terms }{number \ of \ terms} ⇒ 67.5=\frac{x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)}{6}⇒67.5=\frac{6x+15}{6} ⇒ 405=6x+15 ⇒ 390=6x ⇒ x=65\)
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Answers and Explanations: Advanced Algebra and Functions
41- Choice D is correct
You can find the possible values of and b in \((ax+3)(bx-4)\) by using the given equation \(a+b=5\) and finding another equation that relates the variables a and b. Since \((ax+3)(bx-4)=-6x^2+2cx+10\), expand the left side of the equation to obtain
\(abx^2-4ax+3bx-12=-6x^2+2cx+10\)
Since \(ab\) is the coefficient of \(x^2\) on the left side of the equation and 10 is the coefficient of \(x^2\) on the right side of the equation, it must be true that \(ab=-6\)
The coefficient of \(x\) on the left side is \(3b-4a\) and the coefficient of \(x\) in the right side is \(2c\). Then: \(\frac{3b-4a }{2}=c, a+b=5\), then: \(a=5-b\)
Now, plug in the value of a in the equation \(ab=-6\). Then: \(ab=-6→(5-b)b=-6→5b-b^2=-6⇒\) Add \(-5b+b^2\) both sides. Then: \(b^2-5b-6=0\)
Solve for b using the factoring method. \(b^2-5b-6=0→(b-6)(b+1)=0\)
Thus, either \(b=6\) and \(a =-1\), or \(b =-1\) and \(a =6\). If \(b =6\) and \(a =-1\), then:
\(c=\frac{3b-4a}{2}=\frac{3(6)-4(-1) }{2}=11\). If \(b=-1\) and \(a =6\), then, \(c=\frac{3b-4a }{2}=\frac{3(-1)-4(6) }{2}=\frac{27}{2}=13.5→c=13.5\). Therefore, the two possible values for c are 11 and 13.5.
Method 2: Multiplying each side of \(3x+2y=0\) by \((-1)\) gives \(-3x-2y=0\). Then, adding the corresponding side of \(-3x-2y=0\) and \(3x-4y=6\) gives \(-6y=6\). Dividing each side of \(-6y=6\) by \((-6)\) gives \(y=-1\). Finally, substituting \((-1)\) or \(y\) in \(3x+2y=0\), or \(y=\frac{2}{3}\). Therefore, the solution to the given system of equations is \((\frac{2}{3},-1)\).
42- Choice C is correct
If \(f(x)=7x+2(1-x)\), then find \(f(3x)\) by substituting \(3x\) for every \(x\) in the function. This gives: \(f(3x)=7(3x)+2(1-(3x)) ⇒\) It simplifies to: \(f(3x)=7(3x)+2(1-(3x))=21x+2-6x=15x+2\)
43- Choice D is correct
First, find the equation of the line. All lines through the origin are of the form \(y=mx\), so the equation is \(y=\frac{2}{5} x\). Of the given choices, only choice D (15,6), satisfies this equation:
\(y=\frac{2}{5} x→6=\frac{2}{5} (15)=6\)
44- Choice B is correct
\((4n^2-5n+12)-(3n^2-4n)\). Add like terms together: \(4n^2-3n^2=n^2
-5n+4n=-n\). 12 doesn’t have like terms
Combine these terms into one expression to find the answer: \(n^2-n+12\)
45- Choice A is correct
Method 1: Plug the values of x and y provided in the options into both equations.
A. \((\frac{2}{3},-1) 3x+2y=0→3(2/3)+2(-1)=0\)
B. \((5,- 4) 3x+2y=0→3(5)+2(-4)≠0\)
C. \((4,-4) 3x+2y=0→3(4)+2(-4)≠0\)
D. \((\frac{5}{3},-6) 3x+2y=0→3(5/3)+2(-6)≠0\)
Only option A is correct.
46- Choice A is correct
To solve this problem, first recall the equation of a line: \(y=mx+b\)
Where m=slope. \(y=y-\)intercpt
Remember that slope is the rate of change that occurs in a function and that the \(y\)-intercept is the \(y\) value corresponding to \(x=0\).
Since the height of John’s plant is 8 inches tall when he gets it. Time (or \(x\)) is zero. The plant grows 5 inches per year. Therefore, the rate of change of the plant’s height is 5. The \(y\)-intercept represents the starting height of the plant, which is 5 inches.
47- Choice A is correct
Since (1, 2) is a solution to the system of inequalities, substituting 1 for \(x\) and 2 for \(y\) in the given system must result in two true inequalities. After this substitution, \(y>-a-x\) becomes \(-3 < a\), and \(y > x+b\)becomes \(1 > b\). Hence, a is more than -3 and b is less than 1.
Therefore, \(b>a\).
48- Choice C is correct
First, find the slope of the line using the slope formula. \(m=\frac{y_2-y_1}{x_2-x_1 }\)
Substituting in the known information. \((x_1, y_1 )=(1,-5), (x_2, y_2 )= (-3,7)
m=\frac{7-(-5)}{-3-1}=-\frac{12}{4}=-3\)
No,w the slope to find the equation of the line passing through these points. \(y=mx+b\)
Choose one of the points and plug in the values of \(x\) and \(y\) in the equation to solve for b.
Let’s choose point \((1,-5)\). Then: \(y=mx+b→-5=-3(1)+b→-5=-3+b→b=-5+3=-2\), The equation of the line is: \(y=-3x-2\)
Now, plug in the points provided in the choices into the equation of the line.
A. \((2,-3) \ \ \ y=-3(2)-2=-8\) This is NOT true.
B. \((5,-12) \ \ \ y=-3(5)-2=-17\) This is NOT true.
C. \((-4,10) \ \ \ y=-3(-4)-2=10\) This is true.
D. \((3,- 6) \ \ \ y=-3(3)-2=-11\) This is NOT true!
Therefore, the only point from the choices that lies on the line is \((-4,10)\).
49- Choice B is correct
The input value is 5. Then: \(x=-4⇒ f(x)=2x^2+4x+1→f(-4)=2(-4)^2+4(-4)+1=17\)
50- Choice B is correct
To rewrite \(\frac{1}{\frac{1}{x+4}+\frac{1}{x-9}}\), first simplify \(\frac{1}{x+4}+\frac{1}{x-9}\).
\(\frac{1}{x+4}+\frac{1}{x-9}=\frac{1(x-9)}{(x-9)(x+4)}+\frac{1(x+4)}{(x+4)(x-9)}=\frac{(x+4)+(x-9)}{(x+4)(x-9)}\)
Then: \(\frac{1}{\frac{1}{x-9}+\frac{}{x+4}}=\frac{1}{\frac{(x+4)+(x-9)}{(x+4)(x-9)}}=\frac{(x-9)(x+4)}{(x+4)+(x-9)}\). (Remember, \(\frac{1}{\frac{1}{x}}=x)\)
This result is equivalent to the expression in choice B.
51- Choice B is correct
The line passes through the origin, (2,m) and (m,10).
Any two of these points can be used to find the slope of the line. Since the line passes through (0, 0) and (2,m), the slope of the line is equal to \(\frac{m-0}{2-0}=\frac{m}{2}\). Similarly, since the line passes through (0, 0) and (m,10), the slope of the line is equal to \(\frac{10-0}{m-0}=\frac{10}{m}\). Since each expression gives the slope of the same line, it must be true that \(\frac{m}{2}=\frac{10}{m}\)
Using cross multiplication gives, \(\frac{m}{2}=\frac{10}{m}→m^2=20 →m=±\sqrt{20}=±\sqrt{4×5}=±\sqrt{4}×\sqrt{5}=±2\sqrt{5}\)
52- Choice A is correct
The equation of a circle can be written as \((x-h)^2+(y-k)^2=r^2\)
where (h,k) are the coordinates of the center of the circle and r is the radius of the circle. Since the coordinates of the center of the circle are \((1,-2)\), the equation is \((x-1)^2+(y+2)^2=r^2\), where r is the radius. The radius of the circle is the distance from the center \((1,-2)\), to the given endpoint of a radius, \((\frac{1}{3},2)\). By the distance formula, \(r^2=(\frac{1}{3}-1)^2+(2+2)^2=\frac{148}{9}\), Therefore, an equation of the given circle is \((x-1)^2+(y+2)^2=\frac{148}{9}\)
53- Choice C is correct
To solve for cos A, first identify what is known.
The question states that ∆ABC is a right triangle whose \(n∠B=90^\circ\) and sin C \(=\frac{3}{5}\).
It is important to recall that any triangle has a sum of interior angles that equals 180 degrees. Therefore, to calculate cos,A use the complementary angles identity of the trigonometric function. cos A=cos \((90-\)C), Then: cos A=sinC
For complementary angles, the sin of one angle is equal to the \(cos\) of the other angle. cos A \(=\frac{3}{5}\)
54- Choice D is correct
To figure out what the equation of the graph is, first find the vertex. From the graph, we can determine that the vertex is at \((2,-3)\).
We can use vertex form to solve the equation of this graph.
Recall vertex form, \(y=a (x-h)^2+k\), where \(h\) is the \(x\) coordinate of the vertex, and k is the \(y\) coordinate of the vertex. Plugging in our values, you get \(y=a(x-2)^2-3\)
To solve for a, we need to pick a point on the graph and plug it into the equation.
Let’s pick \((4,1), 1=a(4-2)^2-3, 1=a(2)^2-3, 1=4a-3, a=1⇒\) Now the equation is: \(y=(x-2)^2-3\), Let’s expand this, \(y=(x^2-4x+4)-3, y=x^2-4x+1\)
\(y=x^2-4x+1\). The equation in Choice D is the same.
55- Choice B is correct
Multiplying each side of \(\frac{7}{x}=\frac{21}{x+4}\) by \(x(x+4)\) gives \(7(x+4)=21x\),divide two side by 7.
\(x+4=3x or x=2\). Therefore, the value of \(-\frac{x}{3}=-\frac{2}{3}\).
56- Choice D is correct
It is given that \(g(6)=8\). Therefore, to find the value of \(f(g(6))\), then \(f(g(6))=f(8)=12\)
57- Choice C is correct
The area of the triangle is: \(\frac{1}{2}\) AD×BC, and AD is perpendicular to BC. Triangle ADC is a \(30^\circ-60^\circ- 90^\circ\) right triangle. The relationship among all sides of the right triangle \(30^\circ-60^\circ- 90^\circ\) is provided in the following triangle: In this triangle, the opposite side of \(30^\circ\) angle is half of the hypotenuse. And the opposite side of \(60^\circ\) is opposite of \(30^\circ × \sqrt{3}\)
CD = 7, then AD \(= 7 × \sqrt{3}\)
Area of the triangle ABC is: \(\frac{1}{2}\) AD×BC = \(\frac{1}{2} 7\sqrt{3}×14=49\sqrt{3}\)
58- Choice D is correct
It is given that \(g(4)=10\). Therefore, to find the value of \(f(g(4))\), substitute 10 for \(g(4)\).
\(f(g(4))=f(10)=40\).
59. Choice A is correct
The equation of a circle with center (h, k) and radius r is \((x-h)^2+(y-k)^2=r^2\). To put the equation \(x^2+y^2-6x+4y=4\) in this form, complete the square as follows:
\(x^2+y^2-6x+4y=4, (x^2-6x)+(y^2+4y)=4\)
\((x〗^2-6x+9)-9+(y^2+4y+4)-4=4, (x-6)^2+(y+4)^2=17\)
Therefore, the radius of the circle is \(\sqrt{17}\)
60- Choice A is correct
By definition, the tan of any acute angle is equal to the \(cot\) of its complement.
Since angles A and B are complementary angles, therefore:
tan A=cot B
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